Question

Solve ${\sec }^{2}x\mathrm{tany}dy+{\sec }^{2}xdx=0$

Answer 1

We have,

${\sec }^{2}x\tan ydy+{\sec }^{2}xdx=0$

$\Rightarrow {\sec }^{2}x(\tan ydy+dx)=0$

$\Rightarrow \tan ydy+dx=0$

$\Rightarrow \tan ydy=-dx$

$\Rightarrow -dx=\tan ydy$

On integrating, we get,

$\Rightarrow \int 1dx=-\int \tan ydy$

$\Rightarrow x=-\sec y+C$

$\Rightarrow x+\sec y=C$

Hence, this is the answer